MEDICAL GENETICS QUESTION BANK 2007

SINGLE BEST ANSWERQuestions 1 and 2A protein is a dimer that functions as an enzyme, the subunits of which are coded for by an autosomal gene. A nonsense mutation in the third exon of this gene results in premature termination of the gene product.

1. If an individual is heterozygous for this nonsense mutation their biochemical phenotype would show:

A) Complete lack of enzyme activityB) 50% of normal activityC) Normal enzyme activity (100%) because of compensation by the normal gene copyD) Greater than 100% of normal activity because of compensation by the normal gene copyE) A variable level of activity dependent upon the Lyonization of the chromosomes

2. The inheritance pattern for a disease phenotype (0% of normal activity) of the described protein is:

A) Autosomal dominantB) Autosomal recessiveC) Mitochondrial (maternal)D) X-linked dominantE) X-linked recessive

3. The best definition of DNA fingerprinting (also known as DNA typing) is:

A) Use of variable number tandem repeat analysis for the identification of individualsB) Methods of genome sequence analysis for the identification and prosecution of guilty

individualsC) Establishment of a database for statistical estimation of identity between two biological

samplesD) Recognition of phenotypic variation among individualsE) Genetic variation, detected in DNA from a biological sample, used to establish a

biological relationship

4. In meiosis (assuming that no crossing over has occurred) homologous chromosomes segregate at:

A) The first but not the second meiotic divisionB) The second but not the first meiotic divisionC) Both the first and second meiotic divisionsD) Neither the first nor the second meiotic divisionE) Homologous chromosomes segregate in mitosis, not meiosis

5. The “two-hit” hypothesis of the origin of cancer:

A) Was first described in Wilm’s tumorB) Describes how mutation of an oncogene results in cancerC) Explains why tumors are frequently present in both eyes of patients with the familial

form of retinoblastoma and only one eye of patients with sporadic retinoblastomaD) States that loss of one copy of a tumor suppressor gene result in tumor formationE) Explains why breast cancer occurs in 1 of 8 women

6. The purpose of a genetic linkage study is to:

A) Discover the pleiotropic effects of a geneB) Examine the DNA sequence of a mutational siteC) Determine the map location of a disease gene D) Determine if matings are consanguineousE) Compare concordance in MZ and DZ twins

7. If a patient is found to have a chromosomal translocation involving homologous chromosomes, then the risk of having a chromosomally unbalanced offspring is:

A) 2% - 5%B) 25%C) 50%D) 100%E) Dependent upon the age of the mother

8. The best definition of the difference between a Genetic Map and a Physical Map is:

A) A physical map is a collection of markers statistically related and a genetic map consists of pieces of chromosomes or genes.

B) A genetic map is based upon linkage studies and a physical map takes advantage of the genetic engineering of other organisms

C) A genetic map is microscopic and a physical map is macroscopicD) A physical map is useful for loci that are far apart and a genetic map is useful for loci that

are close togetherE) A physical map uses monoclonal antibodies and a genetic map uses protein subunits

9. Which of the common chromosomal aneuploidies does not correlate with advancing maternal age?

A) Down syndrome (trisomy 21) B) Turner syndrome (45,X) C) Edwards syndrome (trisomy 18) D) Patau syndrome (trisomy 13)E) Klinefelter syndrome (47,XXY)

10. Therapeutic abortions can be performed legally in Texas in the first two trimesters of pregnancy (to 24 weeks gestation). Prenatal testing is often performed in such a way that results are available before 24 weeks gestation so an elective termination may still be done if there is an abnormality. Keeping this in mind, fetal echocardiograms are generally not done before 22 weeks gestation because:

A) The heart is not completely formed until that timeB) Amniocentesis is typically done at 15-18 weeks, and those result must be back, firstC) Structural congenital heart disease is never fatalD) The time maximizes the ultrasound resolution of the heartE) The American College of Obstetricians and Gynecologists sets 22 weeks as the date of

this screening test for all women.

Questions 11 and 12:Three years ago, a boy with isolated cleft lip was born to a couple. There is no family history. They were counseled at the time that this was a multifactorial trait and carried an empiric recurrence risk of approximately 3%. The mother is healthy herself and has no identifiable nutritional deficits. The mother recently gave birth to another boy with isolated cleft lip and they seek further counseling.

11. Their recurrence risk for future pregnancies is:

A) Increased approximately doubleB) Unchanged C) Reduced to nearly zero since they have now had their affected child D) Increased dramatically to 50% since both affected infants were boys E) Unable to be determined

12. What is the chance that the affected boys will transmit the isolated cleft lip to their children?

A) The same as for the general populationB) Increased because they are affected and have a positive family historyC) Unable to be determinedD) Depends on the age when the affected boys have childrenE) Less than the general population

13. Mutations of autosomal genes produce disorders that may show either a dominant or recessive inheritance pattern. A dominant pattern is typically exhibited when the normal gene product:

A) Has enzyme activityB) Functions primarily in early embryonic developmentC) Has a limited tissue distributionD) Functions as a structural proteinE) Is an enzyme cofactor

14. During the cell cycle, meaningful clinical cytogenetic data are most readily attainable in:

A) G1 phaseB) G2 phase C) M phase D) S phaseE) G0 phase

Questions 15 and 16: Several members of the family shown below suffer from von Willebrand disease, a clotting disorder that is inherited in an autosomal dominant fashion. The mother is 4 months pregnant and the couple wants to know if the fetus will be affected. DNA was extracted and used for Southern blot analysis. The probe recognized an RFLP (5.3 and 3.1 kb fragments) that is closely linked to the von Willebrand locus.

I.

II.

III. 1 2 3 4

5.3 kb -

3.1 kb -

15. Analysis of the linkage data shows a discrepancy. What is the most likely explanation?

A) Linkage disequilibriumB) Linkage maldistributionC) RecombinationD) Somatic mosaicismE) Imprinting of the von Willebrand gene

16. What is the risk that the fetus, III-4, will be affected with von Willebrand disease?

A) ~0%B) 25%C) 50%D) 75E) ~100%

Questions 17 and 18A woman knows that she is a carrier for Fragile X syndrome. She is interested in knowing what sort of prenatal diagnostic tests are of the most use to her for evaluating her next pregnancy. She is interested in a test that can be done with the highest accuracy for Fragile X as early in the pregnancy as possible. She is not currently pregnant.

17. The procedure of choice for her is most likely:

A) Fetal blood samplingB) Chorionic villus samplingC) Early amniocentesis D) Maternal serum triple screeningE) Targeted ultrasound evaluation of the gender of the fetus

18. The laboratory test(s) of choice is (are):

A) Karyotypic analysis of the X chromosome for the fragile siteB) Alphafetoprotein, human chorionic gonadotropic and unconjugated estriolC) Linkage analysis D) DNA analysis for the fragile X mutationE) Interphase fluorescence-in-situ hybridization studies of the X chromosome

19. A couple’s second child is 3 years old and has just been diagnosed with Tay-Sachs Disease. The chance that the 9-year-old full sibling is a carrier of the disorder is:

A) 1/2B) 1/4C) 2/3D) 1/2 x 1/2E) Dependent on the sex of the proband

20. About 70% of minor malformations involve the:

A) Head and NeckB) Head and ChestC) Head and GenitaliaD) Head and Hands

E) Head and Spine

21. A couple comes to see you because they are related and have heard that they are at an increased risk of having a child with a genetic condition. Both of them are of Ashkenazic Jewish descent. They tell you that the man’s brother has cystic fibrosis, and that several of their mutual relatives have spina bifida. Additional information that you need before giving them any information includes:

A) The degree of relationship between the coupleB) Whether the couple is related through the maternal or paternal sidesC) The severity of the cystic fibrosis in the man's brotherD) Whether the couple is willing to use sperm donation E) The reason that they intermarried

22. The most significant difference between mitosis and meiosis is:

A) In meiosis there are 2 cell divisions without an intervening S phaseB) Only meiosis takes place in the gonadsC) In mitosis, the chromosome number is reduced from 2n to nD) Meiosis ‘freezes’ in stage I in both gendersE) Mitosis occurs in all cells for the entire life span of a person

23. At this point in time, scientists have identified many of the genes responsible for common genetic diseases. When possible, diagnostic studies of such a disease in a family can be accomplished by direct mutation analysis. Occasionally, linkage analysis is useful as a diagnostic tool. Linkage analysis is most frequently useful for diagnostic and counseling purposes when:

A) It is not feasible to test for all the known mutations in the geneB) Not all family members are available for studyC) The carrier or affected status of other family members is unclearD) The precise locus of the gene is not knownE) There is more than one gene known to be involved in the disease phenotype

24. In a molecular diagnostic study using linkage, a result is said to be informative if:

A) The disease gene in question has been sequenced (cloned)B) Detailed phenotypic information is available from the primary relativesC) All of the proband’s primary and secondary relatives are available for studyD) It reveals non-paternityE) The disease and non-disease haplotypes are distinct

25. Oncogenes

A) Are normal DNA repair genesB) Cause most hereditary cancersC) Can result from specific chromosome translocationsD) Are created by mutating tumor suppressor genes

E) Can be activated by losing both copies

26. The first gene found solely by positional cloning without the use of cytogenetic clues was:

A) Huntington Disease B) RetinoblastomaC) Hemophilia AD) Cystic FibrosisE) Polycystic Kidney Disease

27. Cancer is:

A) A disease which uses more than 25% of US health care dollars every yearB) A single diseaseC) Familial in 95% of casesD) A multitude of diseases with the common denominator of gene mutationE) Non-clonal in origin

28. BRCA1 and BRCA2 are the two recently described tumor suppressor genes that are found to predispose to breast cancer. Together they account for about what percentage of breast cancer?

A) 5% in women and 50% in menB) 5%C) 50%D) Depends upon the age of the person in questionE) 85%

29. A child is born and is found to have microcephaly (a small head) and a number of other features that are unusual. You find out that his 36 year old mother was diagnosed as a child with phenylketonuria (PKU), and was on a restricted diet until she was 25 years old. Though you have some test pending, you make a diagnosis of PKU embryopathy. The manifestations are worse than those of the PKU in the mother because:

A) The mother is hemizygous for this X-linked disease and has fortunate LyonizationB) The mother has fewer mutated mitochondriaC) The triplet repeat mutation expanded and caused molecular and clinical anticipationD) The baby’s disease is complicated by a chromosomal aneuploidyE) Elevated phenylalanine levels in the fetus are teratogenic

30. A recently married couple requests counseling because they have just learned that they are first cousins. They are at an increased risk to have children affected with:

A) Multifactorial disordersB) Autosomal dominant disorders

C) Mitochondrial disordersD) Chromosomal disordersE) Sex-Linked disorders

31. Hardy-Weinberg equilibrium is observed under certain conditions. One population phenomenon that is required in order for Hardy-Weinberg to be effective is:

A) SelectionB) Small population sizeC) Decreased fitnessD) Random matingE) Migration

Questions 32-34A 25 year old black woman comes to see you because she is engaged to be married and her fiancé’s brother has sickle cell anemia. She is concerned about having a child with the same condition. She is not currently pregnant. She tells you that she and her fiancé are not related, and that she has sickle cell trait (is a carrier for sickle cell disease).

32. The indication for genetic counseling in this case is:

A) Advanced maternal ageB) Family history of chromosomal abnormalityC) Family history of hereditary disorderD) Exposure to a known teratogenE) Prenatal diagnosis

33. The fiancé is 27 years old and healthy. You tell her that his risk of being a carrier is 2 out of 3. The most appropriate next procedure is:

A) To tell her that prenatal diagnosis can be done on either chorionic villus sampling or amniocentesis

B) To test the fiancé and determine his carrier statusC) To suggest racially-selected and sickle cell-screened sperm donationD) To discuss the medical advances being made in the management of sickle cell diseaseE) To refer her to hematology clinic so that they can address her questions

34. Your consultand wants to know the highest risk possible before any investigation proceeds. You tell her that, since she is a known carrier and her fiancé has a 2 out of 3 chance of being a carrier, their highest risk for having a child with sickle cell disease is:

A) 1 out of 50B) 1 out of 25C) 1 out of 12D) 1 out of 6E) 1 out of 3

Questions 35 and 36:You are asked to see a family in which a genetic condition, polydactyly (too many digits) is present in multiple members. You find out that the grandmother had it, as well as two of her daughters and one son. That son is the father of your consultand. Your consultand is affected, but his sister is not. All affected persons are otherwise healthy.

35. The most likely inheritance pattern for this problem in this family is:

A) Autosomal RecessiveB) Autosomal DominantC) X-linked RecessiveD) X-linked DominantE) Y-linked

36. The risk for the consultand to pass the condition on to his children is:

A) 50%B) 25%C) 100% for daughters and 0% for sonsD) 50% for daughters and 0% for sonsE) 100%

Questions 37 and 38A young woman has a brother and a half-brother who are both profoundly mentally retarded. Further assessment reveals that both have severe hydrocephalus due to stenosis of the aqueduct of Sylvius. In some families, including this one, this kind of hydrocephalus is caused by a mutation in the L1CAM gene and is inherited as an X-linked recessive trait. DNA linkage studies are using a three-allele intragenic marker.

37. What is the carrier status of the consultand (individual marked with an arrow)?

A) She is not a carrierB) She is a heterozygous carrierC) She is a hemizygous carrierD) Cannot be determined

12 8 7

E) Results indicate an abnormality of chromosomal inheritance

38. What is the most likely explanation for the fact that the males have only one marker at this site while the females have two?

A) The lack of a second marker indicates a disease-causing deletionB) Lyonization and inactivation of one X chromosome in the malesC) Recombination between the marker and the disease geneD) The males have only one X chromosomeE) Bad karma

39. Pedigrees are a valuable tool for understanding the inheritance pattern of a disease within a family. At times, it can be difficult to decipher the pattern because of ascertainment problems and the variation in severity of the disease among family members. The most significant clue that a pedigree represents an autosomal dominant inheritance pattern rather than an X-linked pattern is that:

A) There will be a clearly affected person in each generationB) There are equal numbers of affected males and femalesC) All males live long enough to be able to father childrenD) There is male-to-male transmission evident at least onceE) The females are more severely affected than the males

40. It is well documented that chromosomal aneuploidies increase with the increasing maternal

age. Which type of genetic condition increases in frequency with increasing paternal age?

A) Uniparental DisomyB) New Autosomal Dominant DiseaseC) Mitochondrial DiseaseD) Y-Linked DiseasesE) Somatic Mosaicism

41. A couple is known to be at increased risk to have a child with cystic fibrosis because the wife’s sister died of the disease. Their carrier status was checked 6 years ago, confirming the presence of a DF508 mutation in the wife, but failing to find any mutation in the husband. Though their first child was healthy, re-testing of the husband was recommended during their second pregnancy. This recommendation was made because:

A) It is a surreptitious way to confirm paternityB) He is advancing in age and may have developed a new mutationC) Screening of all couples is standard-of-care nationallyD) Advances in knowledge about the CF gene has identified more testable mutationsE) Gene therapy for correction of the carrier state has become available

42. The measure of penetrance of a dominant disease can be thought of mathematically. Complete this equation:

% penetrance =

A) p2 + 2pq + q2B) Expected phenotype distribution/observed phenotype distributionC) # of people with the phenotype/# of people with the genotypeD) MM x MNE) Phenotype in one patient/ complete phenotype

43. In the last century, a group of French Canadians moved to southern Louisiana. One of the members of that community was a gentleman who had an Ashkenazic Jewish heritage, but had recently converted to Catholicism and had joined the group, the Acadians, in their move. He was a carrier for Tay-Sachs disease. He fathered many children and now Tay-Sachs disease had an increased frequency among the Acaidian population in Louisiana. This phenomenon is referred to as:

A) ConsanguinityB) Genetic DriftC) ComplementationD) AllelismE) Founder Effect

44. The newborn screening test for phenylketonuria has a high rate of false positives. There are 15 children who test ‘positive’ by the screening test for every 1 child who has true phenylketonuria. The test parameters are defined this way on purpose because they assure a good ascertainment rate for the children actually affected, and

A) There is a definitive biochemical test to back up the screening test and eliminate the false positives

B) For the children who are carriers of the geneC) There is a definitive mutation test to back up the screening test and eliminate the false

positivesD) For the children who are at risk for neuropathies from aspartamine ingestionE) It allows for less rigid testing procedures

Questions 45-48Apolipoprotein E has received much attention recently because of associations with both cardiovascular disease and Alzheimer's disease. There are 3 common alleles; namely, E2, E3 and E4. Consider the following genotypes and counts:

Genotype Number

E2E2 2 E2E3 64

E2E4 7 E3E3 711 E3E4 169

E4E4 11

Total 964

45. What is the frequency of the E2 allele?

A) .002B) .039C) .076D) .175E) .738

46. What is the frequency of the E3 allele?

A) .858B) .738C) .979D) .711E) .500

Questions 47, 48 Suppose that in another sample of data, one obtained frequencies of E2, E3 and E4 as follows:

f(E2) = .10 f(E3) = .70 f(E4) = .20

47. In a sample of 1000 individuals, how many E3E4 heterozygotes would you expect?

A) 700B) 280C) 140D) 70E) 20

48. Linkage analysis requires that at least one parent be heterozygous at the marker locus. Using the frequencies in problem 52, how many of 1000 individuals are expected to be heterozygous?

A) 140 B) 700C) 460D) 280E) 28

49. The ultimate goal of the Human Genome Project is to:

A) Clone peopleB) Provide information for prenatal manipulation of genes and tailoring of offspringC) Characterize all proteins in the human body down to the quantum levelD) Sequence all the DNA in the human nuclear and mitochondrial DNAE) Develop germ-line gene therapy for the twenty major genetic diseases

50. Mendel’s Law of Independent Assortment states that:

A) Blending of characteristics of the parents does not occurB) Two members of a single gene pair are never found in the same gameteC) How one gene pair segregates is unrelated to how another gene pair segregatesD) Members of a given species preferentially mate with similar partnersE) The presence of the phenotype is independent of the presence of the genotype

51. A woman knows that she is a carrier for cystic fibrosis, an autosomal recessive disease. Her husband’s carrier status has never been tested. If incidence of the disease in the general population is 1 in 1600 live births, what is their baseline risk for having a child with cystic fibrosis?

A) 1/4 B) 1/20C) 1/40D) 1/80E) 1/200

52. By five (5) years of age, how many children may be recognized to have a congenital anomaly?

A) 0.5-2%B) 2-7%C) 7-10%D) 10-17%E) 17-25%

53. The coefficient of inbreeding (F) for a half-sib mating is

A) 1/4B) 1/8C) 1/16D) 1/32E) 1/64

54. Mitochondria are organelles that originated as endosymbiotic prokaryotes. Like chloroplasts in plants, mitochondria have their own genome that is separate from the nuclear genome of the cell. It is possible for a human disease condition to arise as a result of a mutation in the mitochondrial genome. Mitochondrial disease most typically involves the cytochrome oxidase enzymes. Phenotypically these diseases are most likely to effect:

A) Skeletal muscle and central nervous systemB) Cardiac muscle and liverC) Central nervous system and bloodD) Gonads and liverE) Cardiac muscle and blood

55. You have been called to the hospital newborn nursery to see a baby with Down syndrome. The diagnosis is clinically apparent, and you leave the nursery to go talk to the parents about the diagnosis. On the way to the mother’s room, you decide that there is a specific, relatively small, amount of genetic information that you absolutely must give them right now. In your head, you run through various ways of cushioning what you must tell them. The purpose of ‘cushioning’ genetic information about a child’s disease when we give it to families is to:

A) Influence their course of action to what is most medically appropriateB) Delay parental shock until the information is completely toldC) Assuage parental guilt, then you can tell them the important stuff at a later meetingD) Tell the most important information in such a way that it can be understood as fully as

possibleE) Stall for time until the primary care physician has a chance to tell them all the bad news

56. There are many diseases known to be caused by a recently discovered type of mutation referred to as an expanding triplet repeat mutation. These diseases all seem to involve the neuromuscular systems and many are degenerative diseases. Presumably, the expansion of the triplet repeat mutation disrupts the normal transcription/translation of the gene. Relative to the genes themselves on a molecular level, the triplet repeats are found:

A) In the coding regions of the genesB) At different sites depending upon the geneC) In the non-coding regions of the genesD) Only on the maternal chromosomeE) In the transcribed mRNA

57. The father of a 24 year old male consultand has a diagnosis of Huntington disease and is cared for at home. Several paternal relatives are similarly affected, and most are now in long-term care facilities. The consultand and his wife request predictive testing. All appropriate blood samples are obtained in duplicate, and the results indicate that there is nonpaternity; that is, the consultand’s biological father is not the legal and affected father and the consultand has not inherited the Huntington gene. The mother is interviewed separately and gently but does not confirm your genetic interpretation. Which one of the following solutions would best aid the consultand and his wife in planning their future?

A) Respect the autonomy of the mother and tell the patient the molecular testing was not informative

B) Impress upon the mother the importance of divulging the results to her sonC) Respect the autonomy of the mother, say nothing about the nonpaternity, but tell the

patient that the results indicated that he has not inherited the Huntington geneD) Call the couple in and review all the results, including the nonpaternityE) Obtain a court order releasing you from confidentiality to the mother so that you can

reveal all of the results, including the nonpaternity, to the consultand

58. In the space of two years, 5 men have been reported missing in Arizona. Through various means, DNA samples have been found from each of them so that a DNA marker profile could be established. The skeleton of a man has been found recently. DNA fingerprinting has been done in an attempt to identify the skeleton as one of the missing individuals. The gel below represents the results of the testing. The column to the far left is the DNA pattern from the skeleton, and the other 5 columns, A through E, are the DNA patterns from the missing men. The skeleton belongs to which of the missing men?

A) AB) BC) CD) DE) E

59. Anomalies that result from intrinsic abnormalities include:

A) Deformations and DisruptionsB) Disruptions and MalformationsC) Malformations and DysplasiasD) Dysplasias and DisruptionsE) Deformations and Malformations

Questions 60-64Mary is a 27 year old woman who is very involved in the mucopolysaccharidosis (MPS) support

group. Mary’s sister, Jane, died of Hurler syndrome (alpha-L-iduronidase deficiency) at age 8 years. Mary met two very nice men at the support group meeting, Joe and Bob, both of whom are affected with an MPS. Joe has Scheie syndrome (alpha-L-iduronidase deficiency)

SKEL A B C D E

and Bob has Hunter syndrome (iduronate sulfatase deficiency) mild form. All of the defects are verified by enzyme testing. Mary has been dating Joe and Bob.

60. If Mary marries Joe, based on the information presented above, what is the chance that they would have a child affected with an MPS?

A) NegligibleB) 1/8C) 1/4D) 1/3E) 1/2

61. If Mary and Joe do have a son affected with an MPS, the child’s genotype and clinical findings would most likely be an example of which genetic principle(s)?

A) Locus heterogeneityB) Clinical heterogeneityC) Allelic heterogeneityD) Locus and clinical heterogeneityE) Clinical and allelic heterogeneity

62. Mary and Joe’s child who is affected with an MPS would also illustrate which of these genetic principles?

A) Founder effectB) Homozygote for alpha-L-iduronidase mutationC) Compound heterozygote for alpha-L-iduronidase mutationD) Hemizygous for alpha-L-iduronidase mutationE) Genetic heterogeneity

63. If Mary marries Bob, what is the chance that they would have a child affected with an MPS?

A) NegligibleB) 1/8C) 1/4D) 1/3E) ½

64. Hurler syndrome (alpha-L-iduronidase deficiency) which affected Mary’s sister and Hunter syndrome (iduronate sulfatase deficiency) in the severe form can have clinically indistinguishable phenotypes. This is an example of which genetic principle(s)?

A) Locus heterogeneityB) Clinical heterogeneityC) Allelic heterogeneityD) Locus and clinical heterogeneityE) Clinical and allelic heterogeneity

The following information applies to all cases illustrated below in questions 65-70.

A couple has had one affected child with a severe autosomal recessive disease. The disease gene is cloned; however, due to the wide variation in disease causing mutations, it is impossible to perform direct detection of mutation. There is a polymorphism in intron 1 of the disease gene which can be utilized for linkage studies. Please give diagnostic information for the next pregnancy based on the testing that is shown.

65. Concerning the phase and informativeness of the father’s and mother’s chromosomes:

A) Father and mother both phase known and fully informativeB) Father and mother both phase unknown and fully informativeC) Father informative and mother is uninformativeD) Father is uninformative and mother is informativeE) Phase is known in the father but unknown in the mother

66. The current pregnancy is a child who is:

A) Affected with the diseaseB) Unaffected with the disease and a non-carrier of the disease geneC) Unaffected with the disease and a carrier of the father’s copy of the disease geneD) A or BE) B or C

PKb

7

2

67. Concerning the phase and informativeness of the father’s and mother’s chromosomes

A) Father and mother both phase known and fully informativeB) Father and mother both phase unknown and fully informativeC) Father and mother both phase unknown and partially informativeD) Father is informative and mother is uninformativeE) Father is uninformative and mother is informative

PKb

7

2

68. The current pregnancy is a child who is:

A) Affected with the diseaseB) Unaffected with the disease and a non-carrier of the disease geneC) Unaffected with the disease and a carrier of the father’s copy of the disease geneD) A or BE) A or C

69. Concerning the phase and informativeness of the father’s and mother’s chromosomes

A) Father and mother both phase known and fully informativeB) Father and mother both phase unknown and fully informativeC) Father informative and mother is uninformativeD) Father is uninformative and mother is informativeE) Phase is known in the father but unknown in the mother

P

Kb

7

2

70. The current pregnancy is a child who is:

A) Affected with the diseaseB) Unaffected with the disease and a non-carrier of the disease geneC) Unaffected with the disease gene and a carrier of the father’s copy of the disease geneD) A or CE) B or C

Questions 71-80Mr. and Mrs. Jones-Smith comes to your office in their 8th week of pregnancy. They have one 7 year old son with the urea cycle defect of Ornithine Transcarbamylase (OTC) deficiency. He is neurologically devastated from the disease. They are interested in having the current pregnancy tested. They know that OTC deficiency is X-linked and know that it only affects males. Using that logic, they ask about testing the gender of the pregnancy.

71. At this stage of the pregnancy, you tell them that the test for gender identification that can be done to give them the earliest and most specific answer is:

A) Targeted ultrasound looking at the genitaliaB) Early amniocentesis for karyotypingC) Fetal blood sampling for karyotypingD) Chorionic villus sampling for karyotypingE) Routine amniocentesis for karyotyping

72. In the time since the birth of their now 7-year old affected son, new technologies have become available that allow for direct enzyme testing of a fetus. Specifically, it is possible to test the activity of the OTC enzyme in fetal cells and determine whether a fetus is affected. From the point of view of patient and pregnancy management the greatest advantage to this new test is:

A) Differentiation between normal and affected male fetusesB) Increased detection of carrier female fetusesC) Facilitated identification of carrier mothersD) Ability to monitor the effect of the maternal diet on the fetusE) Ability to monitor developmental activation of the enzyme

73. As the couple ponders what you have told them, you wonder whether Mrs. Jones-Smith knows her carrier status. Based on the reasoning of Haldane’s Theorem, you know that her baseline risk of being a carrier for OTC deficiency is:

A) ~0B) 1/4C) 1/3D) 2/3E) ~1

Questions 71-80 (continued)74. If Mrs. Jones-Smith is not a carrier, the chance that this pregnancy will result in an affected

male is:

A) ~0B) 1/4C) 1/3D) 2/3E) ~1

75. You ask the Jones-Smiths if you can have a little more information about their families. They tell you that they each have two siblings. Both of Mrs. Jones-Smith’s siblings have healthy children, while Mr. Jones-Smith has only one nephew. Mrs. Jones-Smith’s mother died last year in a motor vehicle accident and both of Mr. Jones-Smith’s parents are still living and are healthy. No one in the family is on an unusual diet and no one is mentally retarded. As far as the couple knows, they are not blood relatives. The other most significant question that you have to ask is:

A) Whether anyone has had any miscarriages or stillbirthsB) Whether Mrs. Jones-Smith might have been exposed to anything during her previous

pregnancyC) Who the affected child looks like in the familyD) What the ethnic/nationality background of the family isE) Whether there have been any children who died as infants

76. You find out that Mrs. Jones-Smith had a brother who died at a few days of age, reportedly because of an infection. Knowing that the manifestations of OTC deficiency in infants overlap significantly with those of infection, you are suspicious that that child actually died of the genetic disease. Your concerns that Mrs. Jones-Smith is a carrier:

A) Do not change significantlyB) IncreaseC) Decrease slightlyD) Are not affected by the new dataE) Decrease significantly

77. The Jones-Smith’s tell you that they are, indeed, interested in more precise and specific testing if it is available. You do not know yet which test may be most appropriate for them, as you do not know what they plan to do with the information. It is important to ask what their plans may be because:

A) You want to be able to persuade them to terminate because this is a devastating diseaseB) They may not know all of the options available to themC) You prefer to order the most expensive test that their insurance will coverD) You want to confirm that they will do what you would do in this situationE) You do not want them to be saddled with another affected child

Questions 71-80 (continued)78. The couple tells you that they are not interested in terminating a pregnancy, but also do not

wish to have another child who is neurologically devastated. Your recommendation to them is:

A) They meet with some other families to get another perspectiveB) That they reconsider termination as the only logical optionC) That they do not need prenatal testing if they are not going to choose to terminateD) Not to tell anyone, and let the baby die at home if it gets sick in the first few daysE) To deliver at a hospital that is equipped to handle an affected baby if necessary

79. After the appropriate testing, you contact the couple to tell them that the current fetus is, indeed, an affected male. Plans are made to handle the delivery of the baby and his immediate post-partum care. The most important thing about immediate care of the baby is to:

A) Avoid contact with the other family membersB) Feed the baby large amounts of specific amino acidsC) Isolate him from the rest of the nurseryD) Restrict his diet so that he receives a small, controlled amount of proteinE) Remember not to give him the diphtheria vaccine

80. 11 months after birth, the boy, now named William, is healthy, active, and normal in development. He has been carefully monitored and treated by his geneticists and his parents have been fastidious in management of his OTC deficiency. The parents now are seeking a more permanent ‘cure’ for William’s disease. You tell them that the only currently available, long term ‘cure’ is:

A) Liver transplantB) To keep him on his special dietC) Retroviral gene therapyD) Bone marrow transplantE) Weekly infusion of normal gene product

81. Most known genes are in the size range:

A) <1.0 kbB) 10-50 kbC) 100-250 kbD) 250-500 kbE) >1000 kb

82. The term ‘dominant negative allele’ refers to a situation in which:

A) A mutation leads to the production of no protein from one alleleB) The product of the abnormal allele disrupts the function of the product of the normal alleleC) A missense mutation results in the production of unstable protein that is degraded within

the cellD) A retrotransposable element is inserted into a dominant disease geneE) A mutation in the promotor for a gene results in reduced transcription of the gene and

therefore decreased protein product

83. Osteogenesis imperfecta type I is the mildest form of osteogenesis imperfecta because most of the mutations that cause osteogenesis imperfecta type I:

A) Disrupt the folding of the triple helical domain of type I collagen near the carboxyl-terminal domain of the protein

B) Disrupt the cleavage of the amino-terminal domain of the proteinC) Result in no protein being produced from one allele (null allele)D) Result in 3 out of 4 of the type I collagen molecules that are produced to be abnormalE) Prevent the secretion of the alpha-2 chain because of leader-sequence mutation

84. Consent is required in order to enroll people in screening programs of almost any type. The one exception to this rule is newborn screening. All newborns in the United States or babies born to US Military Personnel overseas are subject to screening for phenylketonuria and other genetic diseases. Consent of the parents is not required for this screening to be done. This exception exists because:

A) The public health issue of avoiding the consequences of these diseases takes precedence over family autonomy.

B) Testing on children does not generally require the consent of parentsC) Consent is not required for any genetic screening as long as the results are kept

confidentialD) The lobby for providers of special infant formulas is so strong that congress will not

address the issueE) The courts have mandated that this is an appropriate hospital policy in the face of early

discharges of newborns and their mothers

85. The type of study most useful for determining how much of a disease susceptibility is genetic and how much is environmental is a

A) Linkage studyB) Association studyC) Twin studyD) Segregation analysisE) Prevalence analysis

86. Duchenne and Becker muscular dystrophies both result from mutations in the gene for dystrophin. The diseases differ in onset and severity, but not in inheritance pattern. They should therefore be considered:

A) Pleiotropic disordersB) Allelic disordersC) Variably penetrant disordersD) Closely linked disordersE) Reciprocally expressed disorders

MULTIPLE MATCHINGEach group of items in this section consists of lettered options followed by a set of numbered items. For each item, select the one lettered option that is most closely associated with it. Each lettered option may be selected once, more than once, or not at all.

A) Autosomal Dominant inheritanceB) Autosomal Recessive inheritanceC) X-linked Dominant inheritanceD) X-linked Recessive inheritanceE) New autosomal dominant mutationF) New X-linked mutationG) Chromosomal aneuploidyH) Chromosomal rearrangement

Match each clinical situation with the most likely etiology.

87. Two maternal uncles with a clinical picture similar to the affected male child88. Advanced paternal age89. Advanced maternal age90. Recurrent early pregnancy losses91. Consanguinity

A) ImprintingB) Mitochondrial InheritanceC) Multifactorial InheritanceD) Sex - LimitedE) HemizygousF) HaplotypeG) Mendelian InheritanceH) Autosomal Recessive InheritanceI) Autosomal Dominant InheritanceJ) Sex - LinkedK) X-Linked RecessiveL) X- Linked DominantM) Y-LinkedN) Uniparental disomy

Match the term listed above with its definition:

92. Classic forms of single-gene inheritance 93. Inheritance pattern of a gene on an X or Y chromosome 94. Inactivation of a gene or genes dependent upon the gender of the transmitting parent95. Inheritance of extra-nuclear genetic material96. Parents of an affected child are obligate carriers of the disease gene97. Having only one allele at a locus98. A specific combination of linked alleles

For each item, select the one lettered option that is most closely associated with it. Each lettered option may be selected once, more than once, or not at all.

A) CaucasianB) BlackC) HispanicD) AsianE) Middle EasternF) MediterraneanG) Ashkenazic

Match the genetic disease with the ethnic group in which it is considered to be most common.

99. Alpha-thalassemia100. Beta-Thalassemia101. Gaucher Disease102. Phenylketonuria

A) Missense mutationB) Nonsense mutationC) Frameshift deletionD) In-frame deletionE) Large deletionF) Large duplicationG) RNA splicing mutation

Match the description below with the type of mutation that it represents

103. A patient has Duchenne Muscular Dystrophy (DMD) because exons 38-44 and the intervening intronic DNA are missing.

104. Tay-Sachs disease (hexosaminidase A deficiency) in 18% of Ashkenazi Jews and <1% of non-Ashkenazi is due to a G->C at exon 12 resulting in inclusion of intron 12 in the mRNA

105. A patient has Duchenne Muscular Dystrophy (DMD) because he has two copies of exons 41-45.

106. Most patients in the United States have cystic fibrosis (CF) because of the deletion of 3 base pairs resulting in the loss of a phenylalanine at codon #508.

107. A patient with neurofibromatosis type 1 (NF1) has a change in the first base pair of a glutamine (CAG) to create a STOP (UAG) codon in exon 2. There is no protein produced because of this change.

108. An individual has mild PKU because one copy of the phenylalanine hydroxylase gene has a change at codon #158 of an arginine to a glutamine and the other copy of the phenylalanine hydroxylase gene has a change at codon #261 of arginine to glutamine.

For each item, select the one lettered option that is most closely associated with it. Each lettered option may be selected once, more than once, or not at all.

A) Allelic heterogeneityB) Clinical heterogeneityC) Locus heterogeneityD) Genetic driftE) Founder effectF) Compound heterozygote

Match the term above with its best definition:

109. The production of clinically different phenotypes from mutations in the same gene. 110. The situation in which mutations at two or more distinct loci can produce the same or closely

similar phenotypes.111. The situation in which there are different mutant alleles at the same locus, each capable of

producing an abnormal phenotype.112. The random fluctuations of gene frequencies in populations. 113. An individual with two different mutant alleles at the same locus 114. A high frequency of a mutant gene in a population founded by a small ancestral group when

one or more of the founders was a carrier of the mutant allele.

Questions 115-119

A) Dietary TreatmentB) Vitamin TreatmentC) Specific Drug treatment (not vitamins)D) Replacement of Gene Product

Match the disease below with the most common treatment modality

115. Phenylketonuria116. Gout (Partial HPRT deficiency)117. Hemophilia118. Methylmalonic acidemia119. Wilson Disease

Questions 120-121 There is a particular type of childhood seizure disorder that has been recognized as familial and shows anticipation. A research lab is interested in locating the gene for this disorder. Testing one family with a candidate gene gives the following data:

120. What is the most likely explanation for the gel results?A. Deletion.B. Expanded repeat.C. Sex limitation.D. Heteroplasmy.E. Lyonization.

Answer: A

121. What is the inheritance pattern?A. Autosomal Dominant.B. Autosomal Recessive.C. X-Linked Dominant.D. X-Linked Recessive.E. Mitochondrial.

Answer: A

MEDICAL GENETICS QUESTION BANK ANSWERS - 2005

1 B The question is talking about a protein that functions as an enzyme (in a biochemical pathway) not a structural protein. Consideration of the enzyme as a dimer is a red herring. In general, those persons who are heterozygous for a mutation in a metabolic enzyme coded by an autosomal gene exhibit 50% of the activity of those persons who have two normal copies. (This is may not be applicable for X-linked enzyme genes). Half of the amount of enzyme is typically enough to allow for phenotypic normalcy; however, on the biochemical level, there is half of normal activity in carrier states. In many diseases measurement of enzyme activity provides a way in which to identify the carrier state. Tay Sachs is one example.

2 BThis question is asking for the inheritance pattern of a disease in which the disease state is caused by complete lack of enzyme activity - or homozygosity for a mutated gene. That is autosomal recessive.

3 EThis is the definition of DNA fingerprinting.

4 AHomologous chromosomes segregate at the first meiotic division whether there is crossing over or not. Crossing over may be required for normal disjunction to take place.

5 CIn familial retinoblastoma, susceptible individuals inherit one abnormal copy of the gene. That is the ‘first hit’. It then only requires a single mutational event in each eye to cause cancer. This is the ‘second hit’.

6 CStrictly speaking, the purpose of a genetic linkage study is to determine the relative positions of two loci within the genome. This could be two markers, a marker and a gene, or two genes. In terms of utility for medicine, and an ultimate outcome of the Human Genome Project, the linkage study is used to determine the map location of a disease gene.

7 DThere are two points to make here. First, the patient has been found to have a chromosome abnormality which is a translocation involving homologous chromosomes. The segments cannot be identical; otherwise, the translocation would not be detectable. Second, because both members of the pair of homologues are abnormal, no normal chromosomes will be placed into a gamete; therefore, all gametes will contain an abnormal chromosomal compliment and no normal offspring can result. A translocation involving homologous chromosomes would be a situation such as the presence of an isochromosome 21. As in the case of an isochromosome 21, there is no way to make a normal gamete, so the risk of having a chromosomally unbalanced offspring is 100%.

8 BA genetic map is a theoretical and statistical tool constructed from observation of genes and how frequently there is recombination between them. A physical map is an actual construct of a piece of DNA, usually contained within a bacterial or yeast host.

9 BIt has been found that the nondisjunction event that leads to a 45,X pattern is a paternal meiosis error, so that the ‘missing’ chromosome is the X or Y from the father. As such, this means that the incidence or occurrence of Turner syndrome is independent of maternal age.

10 DThe heart is completely formed by the 9th week gestation. Medicine’s ability to discriminate fetal heart lesions is limited by technology.

11 AAssuming that the diagnosis was correct the first time, then the previous counseling was appropriate; however, with recurrence of the malformation in a subsequent child, the risk for future pregnancies essentially doubles to ~6-8%. At this point reexamination of the primary relatives of the affected boys is indicated in order to confirm that the cleft is an isolated trait, and not the manifestation of a dominant disease. If a syndrome is not found, the risk does not jump any higher than double the previously counseled number. It might also be prudent to review the mother’s diet history as there appears to be a correlation between folic acid intake and clefts.

12 BOnce they are adults, again assuming that the initial diagnosis was correct and this is a simple multifactorial anomaly, their own risk to have affected children is increased over the risk of the general population because they are affected and each have an affected primary relative.

13 DWhen a protein has enzymatic activity or acts as an enzyme cofactor, heterozygosity for a mutation will generally result in 50% activity levels, which typically allows phenotypic normalcy and a recessive inheritance pattern. When a gene product functions as a structural protein, it can have a dominant-negative effect on the product of the normal gene and, so, the disease is inherited in a dominant fashion: it only takes one mutant copy of the gene to disrupt the normal system. The structural or temporal function of the gene is a separate concern and is not directly related to the inheritance pattern.

14 CKaryotypes are obtained during the mitotic (M) phase of the cell cycle. It is the only time in the cell cycle that chromosomes are visible as condensed structures.

15 COf the available answers, the most likely cause of the discrepancy is the presence of a recombination in paternal meiosis causing the mutated gene to segregate in that one sperm with the 5.3 kb marker instead of the 3.1 kb marker. It may help to review some pictures of meiosis, particularly with regard to recombination and movement of alleles between homologous

chromosomes. One thing to be aware of in this sort of situation is non-paternity - the one affected child appears not to have inherited a paternal allele. However, that can be proven/disproven on a molecular level.

16 EWe assume that recombination is a rare event. Given that, the fetus must have inherited the 3.1 kb allele from his father (since the 5.3 kb allele must be maternal). From reviewing the family history of the disease, we note that it is with the 3.1 kb marker that the disease segregates. So, since this is an autosomal dominant condition, the fetus is most likely affected. We say that his risk of being affected is ‘about’ 100% because of the known presence of recombination in the system, and because nothing is ever 100% or 0% in medicine.

17 BOf all those listed, CVS can be done earliest. Fetal blood sampling has the highest accuracy, but cannot be done until late in the second trimester.

18 DFragile X syndrome is caused by a triplet repeat expansion mutation of the FMR1 gene on the X chromosome. The mutation is consistent from person to person, so direct analysis of the mutation is a highly accurate and efficient diagnostic test. Chromosomal analysis may reveal the fragile site on the X chromosome, but the false negative rate is higher than in direct mutation testing. The other tests listed are not useful for the situation presented here.

19 CTay Sachs is typically lethal in childhood. The brother has survived to age 9, so he is extremely unlikely to be homozygous recessive/affected. Therefore, there are three possible genotype combinations left for him, of which 2 of the 3 are heterozygous carrier states. Thus, his risk of being a carrier is 2/3. (Drawing the Punnett square might be helpful.)

20 DThe head/neck and hands are the most detailed parts of the human anatomy. It follows empirically, therefore, that there are relatively more genes involved in the development of those body parts than others. Thus, a deviation from normal development (a minor malformation) is more likely in these structures.

21 AThrough which side the couple is related is not relevant because CF is autosomal recessive, not sex-linked. CF can vary in severity even within a family, so how severely affected the brother is helpful, but not definitive. The reason they intermarried is not relevant to the question. Whether they are willing to use sperm donation may be useful to know at some point in the future because it will impact upon the counseling you can give them and how you can help them, but it is not useful for discussing carrier status and recurrence risk. Knowing to what degree the couple is related will give you an indication of how much of their genome, on average, they share. The more genes they share (the higher their coefficient of consanguinity) the higher their risk of each carrying the same mutated recessive gene. If they are first cousins, they are at higher risk than if they are 3rd cousins once removed.

22 AThe best answer is that there is no intervening DNA synthesis phase between the two stages of meiosis. Meiosis and mitosis both take place in the gonads because those organs are made up of various supporting tissues that undergo mitosis. Mitosis does not occur in all cells for the life of an individual. Cells such as those in the nervous system and skeletal muscle are ‘post-mitotic’ and do not grow further after infancy. (Remember, the reason one ‘bulks-up’ with exercise is because of hypertrophy, not hyperplasia.) The chromosome number is reduced in meiosis, not mitosis. Meiosis does not ‘freeze’ in males, though it in females in stage I at a particular point called dictyotene.

23 AIn such a case as this, the markers have to be linked with little to no recombination to the gene to be useful as a diagnostic test. Some cases of cystic fibrosis are good examples. DNA diagnostics would be used as a first-line test. If, however, DNA diagnostics do not reveal the mutation because the family has something rare for which there is not an ASO, linkage can be useful. Conversely, the 'old test' for Huntington disease was a linkage study with 4% recombination. Not perfect, but the locus of the gene was known that well and so a useful diagnostic study could be done. Once there came to be a better DNA test, the linkage study was dropped. Once in a while, though, there will be a family with a faulty Huntington gene that has a point mutation, or a frame shift for which there is not a good DNA test. At that point, the linkage (with better markers) becomes, once again, useful.

24 EA study is ‘informative’ if the data allows prediction of the phenotype of a currently undiagnosed person - a fetus during pregnancy, or a presymptomatic individual. To know this, it is necessary to be able to distinguish between (or among) the disease genotype and the non-disease genotype(s).

25 COne common example is the chromosomal translocation in chronic myelogenous leukemia (CML) that involves chromosomes 9 and 22. This interrupts two genes and leads to the creation of a new chimeric gene from the fused portions. This chimeric gene can be transcribed into chimeric mRNA which is then translated into a chimeric gene product with abnormal function.

26 DFound in 1989, the CF gene was located by ‘reverse genetics’, or positional cloning. There were no known patients who had visible chromosome abnormalities that contributed to the locating of the gene.

27 DThis is a good, overall definition of cancer.

28 BFamilial breast cancer (regardless of the gender of the patient) makes up only 5% of all cases of breast cancer.

29 EMaternal PKU embryopathy is a ‘classic’ example of the effects of a maternal illness on the development of a fetus. Women who are known to have PKU are advised to remain on or restart their phenylalanine-restricted diet at least 3 months prior to conception. It is also known that phenylalanine is relatively more concentrated in the fetus than in the mother; therefore, the mothers must actually be on a diet that is more strict than usual.

30 AChildren of consanguineous matings are at increased risk of autosomal recessive disorders, which is what one usually thinks of, and of multifactorial disorders. The increased risk of multifactorial disorders reflects the genetic contribution to those conditions and the fact that, as related people share common genes, they are more likely to have a child who inherits two (or more) bad genes that contribute to a multifactorial condition.

31 DThe Hardy-Weinberg law makes some fundamental assumptions (that are not always true of actual human populations): there is random mating, there is a constant mutation rate of the locus under consideration, there is no selection regarding a particular phenotype, the population has no random fluctuation, and there has been no change in population by migration.

32 CThis is the only answer that is appropriate.

33 BTesting of the fiancé can be done easily and efficiently by hemoglobin electrophoresis of a blood sample. If he is not a carrier, then their risk of having an affected child is negligible. If he is a carrier, it would then be appropriate to address issues of sperm donation, prenatal diagnosis, and medical management of affected children either now or at a preconception visit. Referral to hematology clinic can be done if she wishes more detailed explanation of medical management of sickle cell anemia.

34 DHer carrier risk is 1. The fiancé’s carrier risk is 2/3. For any pregnancy, their risk of having an affected child would be 1/4. Thus, the overall risk given the current information is:

1 x 2/3 x 1/4 = 1/635 BThe polydactyly is present to some degree in all generations, travels vertically through the family and exhibits male-to-male transmission. Therefore the inheritance pattern is autosomal dominant.

36 AAny person with an autosomal dominant condition has a 50-50 chance of passing on the gene to a child. The gender of the parent and the child are not relevant in this case.

37 AThe mutation appears to be segregating with the 8kb marker. The consultand did not inherit that marker, so does not carry the mutation.

38 DThis is the typical appearance of a gel electrophoresis for an X-linked condition.

39 DThe question states that there is a pedigree. What feature must be present in the pedigree to rule out XL inheritance as the mode of inheritance? The question is asking about differences between AD and XL diseases, and the correct answer, that there is male-to-male transmission, would successfully distinguish AD and XL, regardless of whether the XL transmission is XLR or XLD.

40 BThis is an observable phenomenon in such conditions as achondroplasia, Apert syndrome, and neurofibromatosis. It is thought to relate to the number of cell divisions that spermatogonia undergo over the life of the male. With an increasing number of cell divisions, there is an increasing chance for error in replication and, thus, mutation.

41 DThe number of mutations now tested is approximately 30. These mutations represent those most common in the Caucasian population. There are a total of >300 mutations identified in the CF gene. Advancing paternal age is not recognized to be a factor in new mutation of recessive disease. (It is seen in new mutation of some dominant diseases.) There is currently not a screening program for CF and there is no currently available gene therapy for the carrier state. What gene therapy is available is for affected persons and is research only.

42 CPenetrance is a function of population: of all people who have the genotype, what percentage shows the phenotype to any degree.

43 EThis is one of the ‘classic’ examples of a founder effect - all occurrences of a given allele within a population can be traced back to a progenitor who gave that allele to a number of children who then formed an inbred, or relatively isolated, population.

44 AScreening tests in general are only useful if there is a more definitive test to support them and distinguish true from false positives.

45 BGenotype Number Allele E2 Allele E3 Allele E4

E2E2 2 4 0 0E2E3 64 64 64 0E2E4 7 7 0 7E2E4 7 7 0 7

E3E3 711 0 1422 0E3E4 169 0 169 169E4E4 11 0 0 11

Total 964 75 1655 187

The frequency of the E2 allele is:

75 / (75+1655+187) = 75 / 1928 = 0.039

46 AThe frequency of the E3 allele is:

1655/ (75+1655+187) = 1655 / 1928 = 0.858

47 BThis is a three allele system and calls for a trinomial expansion (the Hardy-Weinberg equation is a binomial expansion) in which (p), (q), and (r) represent the three alleles. The squared variables represent homozygotes, the mixed variables represent heterozygotes.

(p + q + r)2 = p2 +2pq + q2 + 2qr + r2 + 2pr

if p = f(E2) = 0.10, q = f(E3) = 0.70, and r = f(E4) = 0.20, then

The number of E3E4 heterozygotes would be 2qr in the above equation. 2qr = 2 x 0.7 x 0.2 = 0.28

0.28 x 1000 = 280

48 CThe heterozygotes are represented by the mixed variables. So, the frequency of heterozygotes is:

2pq + 2qr + 2pr = (2 x 0.10 x 0.70) + (2 x 0.70 x 0.20) + (2 x 0.10 x 0.20) = 0.14 + 0.28 + 0.04 = 0.46

0.46 x 1000 = 460

49 DThe Human Genome Project has a specific goal of sequencing the human genome. The other ‘goals’ mentioned are possible outgrowths of that sequencing, but are not part of the directed project. The ethics and practicality of the other ‘goals’ are certainly questionable.

50 CThis is the definition of the Law of Independent Assortment. It should be remembered that this does not apply to linked genes.

51 DThese problems are solved using the Hardy-Weinberg equation. The population disease rate is p2. The carrier rate (the chance of being a heterozygote) is 2pq.

p + q = 1 When p is small, then q =1 for our purposes (p+q)2 = p2 + 2pq + q2

p2=1/1600, so p=1/40. 2pq = 2 x 1/40 x 1 = 1/20 = husband’s carrier risk

The risk of the couple having an affected child is:

Mom’s carrier risk x Dad’s carrier risk x risk of affected if both carriers

1 x 1/20 x 1/4 = 1/80

52 CThis includes those children who have anomalies that are recognized at birth (i.e., cleft lip) as well as those that are discovered/diagnosed later (i.e., horseshoe kidney)

53 BThe coefficient of inbreeding for a half-sib mating is 1/8. Its 1/2 x 1/2 x 1/2 = 1/8. It is necessary to include the common parent in the equation.

54 AThe tissues most frequently affected by mitochondrial disease are those that rely most heavily upon mitochondria for their energy production. Additionally, disease manifests more severely when it affects post-mitotic (non replicating) cells. The liver contains many mitochondria, but has significant turn-over of diseased cells, so is less likely to manifest mitochondrial disease. Skeletal muscle and nerve cells are post-mitotic and are not replaced by functional tissue if they die, so disease manifests more quickly and more severely in these tissues.

55 DIn this case, the ‘important information’ may include what features the baby has, what your diagnosis is, and whether the baby is in immediate medical danger. It should also include a statement that these children are mentally retarded. While the parents should be told that there is nothing that they did/didn’t do to cause this to happen, there will be no way to assuage the guilt that they will feel. In the same manner, there is no way to delay shock and grief. While it is certainly optimum for the parents to be given this information by someone with whom they have established a rapport that is not always possible. If the primary physician is available, s/he and you can present the information together. If not, it is better to tell the family as soon as possible rather than waiting until a later time when the primary physician may be available.

56 BTriplet repeats are not consistently found in any particular place with regard to the genes. The first one found, Fragile X, is far 5' to the coding region, even 5' to the regulatory elements. The myotonic dystrophy gene's repeats are located not in the protein-coding segment of the DNA, but in the noncoding sequence that follows it. The Huntington disease repeats are in the coding region.

57 CAt issue here is the balance of giving enough information without violating the privacy of the patient. Remember that the mother is a patient, too. The first objective should always be to 'DO NO HARM'. By giving the consultand the information he wants, that he is not at risk for Huntington, the obligation to him is fulfilled and his question is answered. By maintaining the confidentiality of the paternity 'problem' the obligation to the mother is also fulfilled. By bringing up the paternity information, which is unfortunate extra information from the test, there is potential harm to all parties involved. It is also to be noted that standardized informed consent forms for DNA mutation testing include a statement that the testing may reveal nonpaternity.

58 DThis is the DNA ‘fingerprint’ that matches that of the skeleton - the banding patterns are identical.

59 CBoth malformations and dysplasias result from abnormalities of the genetic make-up of the cell. Deformations and disruptions result from outside interference of an intrinsically normal process.

60 DHurler and Scheie are both caused by mutations in the gene for alpha-L-iduronidase. They are alleleic. Since Mary is not affected, she has a baseline carrier risk of 2/3. If Mary is a carrier, she is Nr (N=normal r=mutant). To be more specific, she carries one particular mutation, causing Hurler, so we can refer to her genetically as Nr1 Joe is rr for the same gene, but with a different mutation than Mary, so genetically he is r2r2. A child can be Nr2 (normal from Mary, mutant from Joe) or r1r2 (mutant from both Mary and Joe). Genetically, the only kind of gamete that Joe can make will carry the mutant allele (he has no normal allele).The chance that a child will be affected is then the chance that s/he will inherit the mutant allele (r1) from Mary. That is 50%. So

Mary’s carrier risk x risk of Mary passing on mutation = risk of affected child

2/3 x 1/2 = 1/3.

The child would be a compound heterozygote for the mutations, and might be Hurler, Scheie, or Hurler-Scheie.

61 EClinical heterogeneity is the production of clinical different phenotypes from mutations in the same gene. This occurs is situations like Hurler and Scheie which are both mutations of the same gene, but have distinct phenotypes. Allelic heterogeneity is a more molecular. It is the situation in which there are different mutant alleles at the same locus, each capable of producing an abnormal phenotype.

62 CA compound heterozygote is an individual (or a genotype) with two different mutant alleles at the same locus. Presumably, Mary and Joe have different mutant alleles, so that an affected child would inherit alleles of the gene that have two separate, disease causing mutations.

63 AMary’s sister and Bob have distinct diseases. Those two diseases just happen to have phenotypic similarities - they are phenocopies. Each is normal for the other’s mutant gene. Thus, a child would inherit at least one normal allele for each gene (diseases) in question, and would be normal.

64 ALocus heterogeneity is the situation in which mutations at two or more distinct loci can produce the same or closely similar phenotypes.

65 APhase is known in both the father and the mother. The affected child received the 2kb marker from both parents; therefore, in each parent, the 2kb marker segregates with the mutated allele, and the 7kb marker segregates with the normal allele. This information is also fully informative because it can be used to predict the genotype/phenotype of the current pregnancy.

66 BThe current pregnancy has inherited the 7kb marker from each parent. Since it is known that the 7kb marker segregates with the normal allele in both parents, the current pregnancy has not inherited a mutation; therefore, the current pregnancy is unaffected and a non-carrier.

67 CThe affected child inherited one 7kb marker and one 2kb marker. Each of the parents has a 7kb and a 2kb marker. It is impossible to determine whether the affected child inherited the 7kb marker from the mother and the 2kb marker from the father or the opposite. Therefore, phase is unknown in both parents. The information is, however, partially informative because, of the four potential outcomes, it is possible to eliminate one from the prediction.

68 DThe fetus is either identical to the affected brother (Choice A) or the opposite - unaffected non carrier (Choice B). To be a carrier, the fetus would have had to inherit both 7 Kb markers or both 2 kb markers. It is not possible to be more precise and both outcomes must remain in the prediction.

69 DBoth of the father’s alleles segregate with the 2kb marker. It is impossible to distinguish the alleles, therefore, phase cannot be set in the father, and he is uninformative. The mother has one 7kb and one 2kb marker. Since the affected child had to have inherited the 7kb marker from the mother, that must be the marker segregating with the mutant allele. Thus phase can be set in the mother, and she is informative because this information can be used to predict the genotype/phenotype of the next pregnancy.

70 EThe current pregnancy is unaffected because it inherited the mothers normal allele segregating with the 2kb marker. Since the father is uninformative, it is impossible to determine whether the fetus inherited his normal gene - and is therefore an unaffected non-carrier - or his mutant gene - and is therefore an unaffected carrier. Both choices B and C must remain in the prediction because it is not possible to be more precise.

71 DChorionic villus sampling can be done at 10 weeks with results in as early as 72 hours later. Fetal blood sampling eliminates the risk of maternal cell contamination, so is more precise, but cannot be done until late in the second trimester. Targeted ultrasounds have a significant false negative rate, and could not be done until the second trimester. Early amniocentesis is done at 13 weeks. Routine amniocentesis is done at 18-20 weeks.

72 APrior to the advent of this testing, it was not possible to distinguish affected and unaffected males. It was only possible to do sex-selection for female fetuses. This resulted in termination of some normal males.

73 DAccording to Haldane’s theorem, when a male child has a lethal X-linked disease, the chance that the mother is a carrier is 2/3. This has been proven for Duchenne muscular dystrophy. For OTC deficiency, the risk is empirically 75% that the mother is a carrier.

74 AIf this is the case, then the 7 year old affected son represents a new mutation.

75 EFor some reason, people do not think of children who died in infancy as brothers, sisters, aunts, or uncles. Likewise, they are not categorized as stillbirths. Thus, it is necessary to ask separately about children who died as infants. This is particularly important in this family because OTC deficiency can be lethal in infancy if not recognized and treated. The presence in the family of early infant deaths may indicate other OTC affected males.

76 BOTC deficiency is an X-linked recessive disease. If Mrs. Jones-Smith has both an affected brother and an affected son, then she is an obligate carrier and her risk of having another affected child is significant.

77 BThe role of the physician in this instance is to insure that the couple is as fully informed as possible, and to do that it is necessary to know what they are thinking. The remainders of the answers are unethical and/or indicative of directed counseling.

78 EE is the best answer as it does not deny them testing and information, is not coercive, and fulfills their wishes.

79 DThe biochemical abnormality in OTC deficiency results in an inability to clear nitrogen from the system as urea. The primary source of nitrogen in the diet is protein. Proper management of

OTC deficiency patients at all ages is restriction of protein in the diet and administration of medications that function as nitrogen-sinks that provide an alternative method of nitrogen excretion.

80 AContinuance of the special diet is a safe treatment, but is not a cure for the disease. Gene therapy, by retrovirus or infusion of normal gene product, is not possible. Bone marrow transplant would not be useful because the primary defect is in the liver. Liver transplant is not without it’s own risks, but, if successful, would cure the OTC deficiency.

81 BThis is a recognized genetic fact. Keep in mind, though, that this is ‘most known genes’. There are gene that are very small, and at least one (the dystrophin gene) that is 2Mb (2000kb).

82 BThis is the definition of a dominant negative allele. And example of such a situation can be found in diseases that involve structural proteins like Osteogenesis Imperfecta and Marfan syndrome.

83 COsteogenesis Imperfecta (OI) type I is a dominantly inherited disease. There is a dominant negative effect in other types of OI such that the presence of an abnormal gene product disrupts the function of the normal gene product. In OI type I, there is no dominant negative effect because the mutant allele produces no product. The phenotype is the result of a straight decreased production of gene product.

84 AThe state of Texas has also mandated that insurance companies must provide coverage to affected individuals without the use of the ‘pre-existing condition’ clause.

85 CTwin studies are most useful to distinguish genetic and environmental factors. Dizygotic twins raised together have separate genetic makeups, but similar/identical environments. Monozygotic twins raised apart have identical genetic makeups, but different environments. Twin studies allow for control of one or the other component of the question.

86 BThis is allelic heterogeneity. Becker muscular dystrophy (BMD) typically results from an in-frame deletion of the dystrophin gene. Duchenne muscular dystrophy (DMD) typically results from a null or out-of-frame deletion. As a result, the protein folding of dystrophin is less disrupted in BMD and the disease is not as severe.

Multiple Matching

87 DX-linked because the pattern of affected family members is consistent with the inheritance of an X chromosome. Recessive because the mother is not affected.

88 ENew autosomal dominant mutations are recognized to increase in some diseases in correlation to increasing paternal age. Specific diseases in which this is recognized are Achondroplasia and Apert Syndrome.

89 GChromosomal aneuploidy - a difference in number of whole chromosomes as a result of nondisjunction - is recognized to increase in correlation to increasing maternal age.

90 HOf couples experiencing recurrent early pregnancy loss, 10% of the time one or the other parent (more typically the mother) will be a balanced carrier for a chromosomal rearrangement.

91 BIf a couple are genetically related, there is an increased risk that they share common recessive mutations.

92 GMendel described only single-gene inheritance. His work did not include linked genes. He did investigate some two-gene systems, but only in the context of two unlinked genes responsible for distinct parts of the phenotype.

93 JAny gene located on either sex chromosome is said to be sex-linked.

94 AThis is the definition of imprinting.

95 BMitochondria have a genome separate from that in the cellular nucleus. (The same is true for chloroplasts in plants.) The genes in the mitochondrial genome code for some of the components of the electron transport chain, and mutations in these genes are known to cause human disease.

96 HIn a situation in which a child has a known recessive disease, the parents will be unaffected, but must both be carriers of the disease.

97 EThis is the definition of hemizygosity. It is most frequently used to describe genes on the X chromosome in males, but this is not an exclusive use.

98 FThis is the definition of a haplotype.

99 D, 100 F, 101 G, 102 A

This is very basic medical information regarding genetic diseases common to large ethnic populations. For further information, please refer to the population genetics portion of your required reading.

103 EThis is an absence of a large section of the dystrophin gene. It is a large deletion.

104 GThe mutation changes the sequence of the mRNA such that appropriate processing of it does not take place and an intron is included in the final mRNA product. Normally the introns are spliced out. Therefore, this is a splicing mutation.

105 F

106 D

107 B

108 A

109 B This is the definition of clinical heterogeneity. It is different from allelic heterogeneity because it is identified on a clinical rather than a molecular level. In some conditions, there can be clinical heterogeneity among persons who all have the same mutation. Clinical heterogeneity does not automatically imply that there is allelic heterogeneity.

110 CThis is the definition of locus heterogeneity.

111 AThis is the definition of allelic heterogeneity. It is a more molecular definition.

112 DThis is the definition of genetic drift.

113 FThis is the definition of a compound heterozygote.

114 EThis is the definition of founder effect.

115 APhenylketonuria (PKU) treated by dietary restriction of protein (specifically the amino acid phenylalanine) in the diet. There are rare cases of PKU that are responsive to vitamin therapy, but those are not the norm.

116 CGout is treated by administration of allopurinol, a drug which blocks an enzymatic step that would otherwise lead to an accumulation of uric acid.

117 DHemophilia is treated by administration of partially purified coagulation factor VIII, which is the normal version of the genetically abnormal gene product.

118 BMethylmalonic acidemia is treated by administration of B12.

119 C Wilson disease is treated by administration of penicillamine. This acts as a chelator which increases excretion of stored copper.

120 A

121 A